How do you factor \[{x^2} - 5x + 6 = 0\]?

Hint: The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. The roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. In factorization if it’s difficult to split the middle terms we use quadratic formula or Sridhar’s formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step by step answer:
The degree of the equation \[{x^2} - 5x + 6 = 0\] is 2, so the number of roots of the given equation is 2.
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\].
We get \[a = 1\], \[b = - 5\]and \[c = 6\].
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\].
In the given question, we have to find the value of \[{b_1}\] and \[{b_2}\] by hit and trial method such that \[{b_1} \times {b_2} = 6\] and \[{b_1} + {b_2} = - 5\].
That Is we have \[{b_1} = - 2\] and \[{b_1} = - 3\] because it satisfies the conditions \[{b_1} \times {b_2} = 6\] and \[{b_1} + {b_2} = - 5\].
Then we have,
\[ \Rightarrow {x^2} - 2x - 3x + 6 = 0\]
Taking ‘x’ common in the first two terms and taking \[ - 3\]from the remaining two terms.
\[x(x - 2) - 3(x - 2) = 0\]
Again taking \[(x - 2)\] common we have,
\[ \Rightarrow (x - 3)(x - 2) = 0\]

Thus the factors are \[(x - 3)\] and \[(x - 2)\].

Note: We can also find the roots of the above quadratic equation by substituting the obtained factors to zero.
\[ \Rightarrow (x - 3) = 0\] and \[(x - 2) = 0\]
\[ \Rightarrow x = 3\] and \[x = 2\] are the roots.
In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis, that is the roots are simply the x-intercepts.